# Laplace question continued (partial fractions)

Last night I attempted and successfully finished (with the help of stackexchange) the first part to this question on laplace transformations:

https://math.stackexchange.com/questions/468596/laplace-question-help-needed

The second part to this question is:

Here are my working/calculations so far:

I know I am going wrong somewhere but I don't know where. When solving the last two equations (shown) simultaneously I am getting different answers for C.

Any help on this is really appreciated! thank you :)

• Hint: Use $Bs$ as the numerator instead of $B$. In fact I would have tried the more general form $\frac{As+B}{s^2}+\frac{Cs+D}{s^2+1}$ for RHS, if I remember correctly. – Macavity Aug 16 '13 at 7:32
• what about for the third denominator? (s^2+1) – Anona anon Aug 16 '13 at 8:02
• You have not multiplied out your fractions consistently - the first on the left seems to have been multiplied by $s^2(s^2+1)^2$, the second by $s^2(s^2+1)$ - not sure why you need $(s^2+1)^2$. Anyway, you can substitute $s=\pm 1$ in the original equation without multiplying through. If you want to eliminate terms use $s=\pm i$ – Mark Bennet Aug 16 '13 at 8:19

You appear to be working with (using a conventional general form of partial fractions) $$\frac{1}{s^2(s^2+1)}+\frac{s-2}{s^2+1}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+1}$$ Multiply through by $s^2(s^2+1)$ to clear fractions$$1+s^2(s-2)=(As+B)(s^2+1)+(Cs+D)s^2$$ We then set $s=0$ to obtain $B=1$ and setting $s=i$ gives $$3-i=-Ci-D$$ We can then equate the coefficients of $s^3$ on each side to obtain $A$.

Another route would be to work only with the first term on the left (since the second term on the left is in the correct form anyway), and note that this is an expression in $t=s^2$ with $$\frac 1{t(t+1)}=\frac At+\frac B{t+1}=\frac 1t-\frac 1{t+1}$$ where we use conventional partial fractions technique (or prior knowledge of a common form) to obtain the expression.

Using the first method we obtain $B=1, C=1, D=-3$ from the given equations. Equating coefficients of $s^3$ gives $s^3=As^3+Cs^3$ and since $C=1$ we have $A=0$ giving $$\frac{1}{s^2}+\frac{s-3}{s^2+1}=\frac{1}{s^2}+\frac{s}{s^2+1}-\frac{3}{s^2+1}$$

Using the second method ($t=s^2$) we get $$\frac 1t-\frac 1{t+1}+\frac{s-2}{s^2+1}=\frac 1{s^2}-\frac 1{s^2+1}+\frac{s-2}{s^2+1}=\frac1{s^2}+\frac s{s^2+1}-\frac3{s^2+1}$$

• Quick question - why are you setting s to a letter(i)? What is the reason behind this and in what instances is this supposed to be done? (I'm asking because I haven't done that before in any partial fraction questions that I have done). – Anona anon Aug 16 '13 at 11:29
• @Anonaanon I am using $i=\sqrt{-1}$ so that $i^2=-1$ and $i^2+1=0$, which means that the $s^2+1$ terms all go to zero. – Mark Bennet Aug 16 '13 at 11:37
• Okay thank you for that. Also how can B=1 when in the question A=1 and B=s? (the given result) – Anona anon Aug 16 '13 at 13:41
• @Anonaanon I've used different letters from you, but I've filled out the solution a bit at the end, so you can see where I am going. – Mark Bennet Aug 16 '13 at 13:58

multiply both sides by $$s^2(s^2+1)^2$$ and then do it. and it should be $$\frac{A}{s} + \frac{B}{s^2} + \frac{Cx+D}{s^2+1} + \frac{Ex+F}{(s^2+1)^2}$$ Sorry i dont know how to write in math format.

• @safwanashfaq: Why do you have the $Ex + F$ term? – Amzoti Aug 16 '13 at 12:59
• because when we have power of some irreducible factor then it appears as many times as its power it. – safwanashfaq Aug 16 '13 at 15:34